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BigorU [14]
3 years ago
15

A charge of 50 µC is pushed by a force of 25 µN a distance of 15 m in an electric field. What is the electric potential differen

ce?
Physics
1 answer:
TiliK225 [7]3 years ago
8 0
7.5 is the correct answer
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What is the value of a conversion factor ratio?<br>A.1<br>B.3<br>C.10<br>D.12
Alecsey [184]
A is the correct answer :D

4 0
4 years ago
What is acceleration?
bonufazy [111]
I'm pretty sure it's A.
4 0
3 years ago
4. Which of the following would be a good reference point to describe the motion of a dog?
saul85 [17]

ANOTHER RUNNING DOG

Explanation:

In the given question it is to find a suitable reference point to describe  the motion of dog. Here I could suggest that it is better to compare the dog with  another running dog to create the relative speed difference to get a reliable motion variation.

Because the motion of dog is in the linear with respect to the another dog and to the acceleration produced by the dog in the required interval is easy to calculate with respect to  another dog which is already in motion.

Hence, I suggest that Motion of dog can be analysed better by analyse the motion variation of dog with  another dog running.

4 0
3 years ago
The mean distance of an asteroid from the Sun is 2.98 times that of Earth from the Sun. From Kepler's law of periods, calculate
lutik1710 [3]

Answer:

The asteroid requires 5.14 years to make one revolution around the Sun.

Explanation:

Kepler's third law establishes that the square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit:

T^{2} = a^{3} (1)

Where T is the period of revolution and a is the semi-major axis.

In the other hand, the distance between the Earth and the Sun has a value of 1.50x10^{8} Km. That value can be known as well as an astronomical unit (1AU).

But 1 year is equivalent to 1 AU according with Kepler's third law, since 1 year is the orbital period of the Earth.

For the special case of the asteroid the distance will be:

a = 2.98(1.50x10^{8}Km)

a = 4.47x10^{8}Km

That distance will be expressed in terms of astronomical units:

4.47x10^{8}Km.\frac{1AU}{1.50x10^{8}Km} ⇒ 2.98AU

Finally, from equation 1 the period T can be isolated:

T = \sqrt{a^{3}}

T = \sqrt{(2.98)^{3}}  

T = \sqrt{26.463592}

T = 5.14AU

Then, the period can be expressed in years:

5.14AU.\frac{1yr}{1AU} ⇒ 5.14 yr

T = 5.14 yr

Hence, the asteroid requires 5.14 years to make one revolution around the Sun.

8 0
3 years ago
Which is an example of a way that
geniusboy [140]

Answer:

C Thy make the force exert all at once

4 0
2 years ago
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